Given a matrix of order NxN, Find the minimum number of steps to convert the given matrix into the Upper Hessenberg matrix. In each step, the only operation allowed is to decrease or increase any element value by 1.
Examples:
Input : N=3
1 2 8
1 3 4
2 3 4
Output :2
Decrease the element a[2][0] 2 times.
Now the matrix is upper hessenberg
Input : N=4
1 2 2 3
1 3 4 2
3 3 4 2
-1 0 1 4
Output :4
Approach:
- For a matrix to be an Upper Hessenberg matrix all of its elements below the sub-diagonal must be equal to zero, i.e Aij = 0 for all i > j+1..
- The minimum number of steps required to convert a given matrix in the upper Hessenberg matrix is equal to the sum of the absolute values of all Aij for all i > j + 1.
- The modulus value of the element is taken into account because both the increase and decrease of the element count as a single step.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>#define N 4using namespace std;// Function to count steps in// conversion of matrix into upper// Hessenberg matrixint stepsRequired(int arr[][N]){ int result = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // if element is below sub-diagonal // add abs(element) into result if (i > j + 1) result += abs(arr[i][j]); } } return result;}// Driver codeint main(){ int arr[N][N] = { 1, 2, 3, 4, 3, 1, 0, 3, 3, 2, 1, 3, -3, 4, 2, 1 }; // Function call cout << stepsRequired(arr); return 0;} |
Java
// Java implementation of above approachclass GFG{ static int N = 4; // Function to count steps in // conversion of matrix into upper // Hessenberg matrix static int stepsRequired(int arr[][]) { int result = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // if element is below sub-diagonal // add abs(element) into result if (i > j + 1) result += Math.abs(arr[i][j]); } } return result; } // Driver code public static void main (String[] args) { int arr [][] = new int [][] {{1, 2, 3, 4}, {3, 1, 0, 3}, {3, 2, 1, 3}, {-3, 4, 2, 1 }}; // Function call System.out.println(stepsRequired(arr)); }}// This code is contributed by ihritik |
Python3
# Python3 implementation of above approachN = 4;# Function to count steps in# conversion of matrix into upper# Hessenberg matrixdef stepsRequired(arr): result = 0; for i in range(N): for j in range(N): # if element is below sub-diagonal # add abs(element) into result if (i > j + 1): result += abs(arr[i][j]); return result;# Driver codearr = [[1, 2, 3, 4], [3, 1, 0, 3], [3, 2, 1, 3], [-3, 4, 2, 1]];# Function callprint(stepsRequired(arr));# This code is contributed by Rajput-Ji |
C#
// C# implementation of above approachusing System;class GFG{ static int N = 4; // Function to count steps in // conversion of matrix into upper // Hessenberg matrix static int stepsRequired(int [, ] arr) { int result = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { // if element is below sub-diagonal // add abs(element) into result if (i > j + 1) result += Math.Abs(arr[i, j]); } } return result; } // Driver code public static void Main () { int [ , ] arr = new int [, ] { {1, 2, 3, 4}, {3, 1, 0, 3}, {3, 2, 1, 3}, {-3, 4, 2, 1}}; // Function call Console.WriteLine(stepsRequired(arr)); }}// This code is contributed by ihritik |
Javascript
<script>// Java script implementation of above approachlet N = 4; // Function to count steps in // conversion of matrix into upper // Hessenberg matrix function stepsRequired(arr) { let result = 0; for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { // if element is below sub-diagonal // add abs(element) into result if (i > j + 1) result += Math.abs(arr[i][j]); } } return result; } // Driver code let arr =[[1, 2, 3, 4], [3, 1, 0, 3], [3, 2, 1, 3], [-3, 4, 2, 1 ]]; // Function call document.write(stepsRequired(arr));// This code is contributed by mohan pavan</script> |
Output:
10
Time complexity : O(N*N)
Auxiliary space: O(1) because it is using constant space for variables
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